What is the determinant of Bernoulli matrix?

Question

Let $n$ be an even integer, what is the determinant of

$$D_{n+1}=\begin{vmatrix} 1&1&\cdots&1&1\\ 2&2^2&\cdots&2^n&2^{n+1}\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&n^2&\cdots&n^n&n^{n+1}\\ \frac{n}{2}&\frac{n^2}{3}&\cdots&\frac{n^n}{n+1}&\frac{n^{n+1}}{n+2}\\ \end{vmatrix}.$$ At first, we can compute that $D_3=D_5=0$, so there is a high possibility that $D_{n+1}=0$. The last row of the determinant remind me the Bernoulli identity \begin{align*} 1+2+\cdots+n &= \frac{n^2}{2}+\frac{n}{2} \\ 1^2+2^2+\cdots+n^2 &=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\\ 1^3+2^3+\cdots+n^3 &=\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4}\\ 1^4+2^4+\cdots+n^4 &=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}\\ 1^k+2^k+\cdots+n^k &=\frac{n^{k+1}}{k+1}+\frac{n^k}{2}+{\color{red}\frac1{k+1}\left( \binom{k+1} 2B_2n^{k-1}+\cdots+\binom{k+1}kB_kn\right)} \end{align*} Then we change the determinant to $$D_{n+1}=\frac{-1}{n}\begin{vmatrix} 1&1&\cdots&1&1\\ 2&2^2&\cdots&2^n&2^{n+1}\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&n^2&\cdots&n^n&n^{n+1}\\ -\frac{n^2}{2}&-\frac{n^3}{3}&\cdots&-\frac{n^{n+1}}{n+1}&\frac{-n^{n+2}}{n+2}\\ \end{vmatrix}.$$ Then add the first $n$ line to the last line to obtain $$D_{n+1}=\frac{-1}{n}\begin{vmatrix} 1&1&\cdots&1&1\\ 2&2^2&\cdots&2^n&2^{n+1}\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&n^2&\cdots&n^n&n^{n+1}\\ \frac{n}{2}&\frac{n^2}{2}+\frac{n}{6}&\cdots&\frac{n^{n}}{2}+\frac{n^n}{12}{\color{red}+\cdots}&\frac{n^{n+1}}{2} +\frac{n^{n+1}}{12}+\frac{n^n}{12}{\color{red}+\cdots}\\ \end{vmatrix}.$$ Now Subtract $n$ times the first column from the second column, $n^2$ times the first column from the third column, and $n^3$ times the first column from the fourth column and so on we obtain $$D_{n+1}=\frac{-1}{n}\begin{vmatrix} 1&\ast&\cdots&\ast&\ast\\ 2&\ast&\cdots&\ast&\ast\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&0&\cdots&0&0\\ \frac{n}{2}&\frac{n}{6}&\cdots&\frac{n^n}{12}{\color{red}+\cdots}&\frac{n^{n+1}}{12}+\frac{n^n}{12}{\color{red}+\cdots}\\ \end{vmatrix}.$$ Then add the $-1/2$ times of the penultimate column to the last column we have that $$D_{n+1}=\frac{-1}{n}\begin{vmatrix} 1&\ast&\cdots&\ast&\ast\\ 2&\ast&\cdots&\ast&\ast\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&0&\cdots&0&0\\ 0&\frac{n}{6}&\cdots&\frac{n^n}{12}{\color{red}+\cdots}&\frac{n^{n+1}}{12}+\frac{n^n}{12}{\color{red}+\cdots}\\ \end{vmatrix}.$$ Then use the $n/6$ in the last line to eliminate all other elements in the last line we have that $$D_{n+1}=\frac{-1}{n}\begin{vmatrix} 1&\ast&\cdots&\ast&\ast\\ 2&\ast&\cdots&\ast&\ast\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&0&\cdots&0&0\\ 0&\frac{n}{6}&\cdots&0&0\\ \end{vmatrix}.$$ At this point, I don't know how to proceed. Can anyone help me？

Answer 1

Formulated in Mathematica, for the first rows

     M[n_] := Outer[(#1^#2 &), Range[n], Range[n + 1]]   

and the last row appended

     MM[n_] := Append[M[n],  n^Range[1, n + 1]/ Range[2, n + 2] ]  

one finds a linear combination of the first rows equating the last row

Solve[{a, b, c, d, e, f} . M[6] == MM[6][[-1]], {a, b, c, d, e, f}][[1]] 

  {{a -> 216/105, b -> 27/840, c -> 272/105, d -> 27/840, e -> 216/105,
  f -> 41/840}}

or commmon denominator removed

  {a -> 216, b -> 27, c -> 272, d -> 27, e -> 216, f -> 41}

This indicates, that trivial line by line reductions will not work the easy way.

Have a look at the [Vandermonde matrix][1]

Answer 2

Let $$f(x)=\begin{vmatrix} 1&1&\cdots&1&1\\ 2&2^2&\cdots&2^n&2^{n+1}\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&n^2&\cdots&n^n&n^{n+1}\\ \frac{x}{2}&\frac{x^2}{3}&\cdots&\frac{x^n}{n+1}&\frac{x^{n+1}}{n+2}\\ \end{vmatrix},$$ then $D_{n+1}=f(n)$ and we have that \begin{eqnarray*} % \nonumber to remove numbering (before each equation) (xf(x))'&=&\begin{vmatrix} 1&1&\cdots&1&1\\ 2&2^2&\cdots&2^n&2^{n+1}\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ n&n^2&\cdots&n^n&n^{n+1}\\ x&x^2&\cdots&x^n&x^{n+1}\\ \end{vmatrix}=n!x\begin{vmatrix} 1&1&\cdots&1&1\\ 1&2&\cdots&2^{n-1}&2^{n}\\ \vdots&\vdots&\cdots&\vdots&\vdots\\ 1&n&\cdots&n^{n-1}&n^{n}\\ 1&x&\cdots&x^{n-1}&x^{n}\\ \end{vmatrix} \\ &=&n!x(x-1)(x-2)\cdots(x-n)\prod_{1\leq i<j\leq n}(j-i)=\left(\prod_{k=1}^nk!\right)x(x-1)(x-2)\cdots(x-n). \end{eqnarray*} Hence $$xf(x)=\int_0^x(tf(t))'\,dt=\left(\prod_{k=1}^nk!\right)\int_0^xt(t-1)(t-2)\cdots(t-n)\,dt.$$ Since $n$ is even, we have that $$nf(n)=\left(\prod_{k=1}^nk!\right)\int_0^nt(t-1)(t-2)\cdots(t-n)\,dt=0.$$ Here we use the fact that function $$g(x):=x(x-1)(x-2)\cdots(x-n)$$ is an odd function with respect to $x=n/2$. For example, $f(x)=x^3$ or $\sin x$ is an odd function with respect to $x=0$ in the usuall sense.